Solve (x-3)^2/4=x

$$\frac { ( x - 3 ) ^ { 2 } } { 4 } = x$$

Solve for x

$$x=1$$

$$x=9$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\frac{x^{2}-6x+9}{4}=x$$

Divide each term of x^{2}-6x+9 by 4 to get \frac{1}{4}x^{2}-\frac{3}{2}x+\frac{9}{4}.

$$\frac{1}{4}x^{2}-\frac{3}{2}x+\frac{9}{4}=x$$

Subtract x from both sides.

$$\frac{1}{4}x^{2}-\frac{3}{2}x+\frac{9}{4}-x=0$$

Combine -\frac{3}{2}x and -x to get -\frac{5}{2}x.

$$\frac{1}{4}x^{2}-\frac{5}{2}x+\frac{9}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -\frac{5}{2} for b, and \frac{9}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\left(-\frac{5}{2}\right)^{2}-4\times \frac{1}{4}\times \frac{9}{4}}}{2\times \frac{1}{4}}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-4\times \frac{1}{4}\times \frac{9}{4}}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-\frac{9}{4}}}{2\times \frac{1}{4}}$$

Multiply -1 times \frac{9}{4}.

$$x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25-9}{4}}}{2\times \frac{1}{4}}$$

Add \frac{25}{4} to -\frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{4}}{2\times \frac{1}{4}}$$

Take the square root of 4.

$$x=\frac{-\left(-\frac{5}{2}\right)±2}{2\times \frac{1}{4}}$$

The opposite of -\frac{5}{2} is \frac{5}{2}.

$$x=\frac{\frac{5}{2}±2}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$x=\frac{\frac{5}{2}±2}{\frac{1}{2}}$$

Now solve the equation x=\frac{\frac{5}{2}±2}{\frac{1}{2}} when ± is plus. Add \frac{5}{2} to 2.

$$x=\frac{\frac{9}{2}}{\frac{1}{2}}$$

Divide \frac{9}{2} by \frac{1}{2} by multiplying \frac{9}{2} by the reciprocal of \frac{1}{2}.

$$x=9$$

Now solve the equation x=\frac{\frac{5}{2}±2}{\frac{1}{2}} when ± is minus. Subtract 2 from \frac{5}{2}.

$$x=\frac{\frac{1}{2}}{\frac{1}{2}}$$

Divide \frac{1}{2} by \frac{1}{2} by multiplying \frac{1}{2} by the reciprocal of \frac{1}{2}.

$$x=1$$

The equation is now solved.

$$x=9 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\frac{x^{2}-6x+9}{4}=x$$

Divide each term of x^{2}-6x+9 by 4 to get \frac{1}{4}x^{2}-\frac{3}{2}x+\frac{9}{4}.

$$\frac{1}{4}x^{2}-\frac{3}{2}x+\frac{9}{4}=x$$

Subtract x from both sides.

$$\frac{1}{4}x^{2}-\frac{3}{2}x+\frac{9}{4}-x=0$$

Combine -\frac{3}{2}x and -x to get -\frac{5}{2}x.

$$\frac{1}{4}x^{2}-\frac{5}{2}x+\frac{9}{4}=0$$

Subtract \frac{9}{4} from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{4}x^{2}-\frac{5}{2}x=-\frac{9}{4}$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}x^{2}-\frac{5}{2}x}{\frac{1}{4}}=-\frac{\frac{9}{4}}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$x^{2}+\left(-\frac{\frac{5}{2}}{\frac{1}{4}}\right)x=-\frac{\frac{9}{4}}{\frac{1}{4}}$$

Divide -\frac{5}{2} by \frac{1}{4} by multiplying -\frac{5}{2} by the reciprocal of \frac{1}{4}.

$$x^{2}-10x=-\frac{\frac{9}{4}}{\frac{1}{4}}$$

Divide -\frac{9}{4} by \frac{1}{4} by multiplying -\frac{9}{4} by the reciprocal of \frac{1}{4}.

$$x^{2}-10x=-9$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-9+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-9+25$$

Add -9 to 25.

$$x^{2}-10x+25=16$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-5=4 x-5=-4$$

Add 5 to both sides of the equation.

$$x=9 x=1$$